Question

Prove that, if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Answer 1

Given: Let a ΔABC in which a line DE parallel to BC intersects AB at D and AC at E.

To Prove: DE divides the two sides in the same ratio.

`"AD"/"DB" = "AE"/"EC"`

Construction: Join BE, CD and draw EF ⊥ AB and DG ⊥ AC.

Proof: Here,

Area of triangle = `1/2` × base × height

Area of ΔADE = `1/2` × AD × EF      

or

Area of ΔADE = `1/2` × AE × DG     

Similarly,

Area of ΔBDE = `1/2` × DB × EF      

Area of ΔDEC = `1/2` × EC × DG     

`"ar(ΔADE)"/"ar(ΔBDE)" = (1/2 × "AD" × "EF")/(1/2 × "DB" × "EF")`

`"ar(ΔADE)"/"ar(ΔBDE)" = "AD"/"DB"`     ...(1)

From (2) and (4),

`"ar(ΔADE)"/"ar(ΔDEC)" = (1/2 × "AE" × "DG")/(1/2 × "EC" × "DG")`

`"ar(ΔADE)"/"ar(ΔDEC)" = "AE"/"EC"`     ...(2)

Since, ΔBDE and ΔDEC lie between the same parallel DE and BC and on the same base DE.

∴ ar(ΔBDE) = ar(ΔDEC)       ...(3)

From (1), (2) and (3), we get,

`"AD"/"BD" ="AE"/"EC"`

Hence proved.